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Thepotemich [5.8K]
3 years ago
5

A truck and a scooter moving with the same velocity colide. Why does the scooter fly off?

Physics
1 answer:
iragen [17]3 years ago
6 0
We know, Momentum is the product of mass and velocity. As velocity is equal, momentum of Truck will be greater as mass of Truck is greater than the scooter. As momentum is greater from truck side, it will transfer the same amount to the scooter, so it flies away

Hope this helps!
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Can someone pls answer number 8 for me pls
fomenos

Answer:

1.33 m

Explanation:

Tom weighs 600 N and is 2 meters from the pivot.  So his moment is:

τ = Fr

τ = (600 N) (2 m)

τ = 1200 Nm

To balance the see-saw, Jeremy must create an equal and opposite moment.

τ = Fr

1200 Nm = (900 N) r

r = 1.33 m

So Jeremy must sit 1.33 m from the pivot on the other side.

4 0
3 years ago
(d) If η = 40% and TH = 427°C, what is TC, in °C?
Brrunno [24]

Answer:

T_C=256.2^{\circ}C

Explanation:

Given that,

Efficiency of heat engine, \eta=40\%=0.4

Temperature of hot source, T_H=427^{\circ}C

We need to find the temperature of cold sink i.e. T_C. The efficiency of heat engine is given by :

\eta=1-\dfrac{T_C}{T_H}

T_C=(1-\eta)T_H

T_C=(1-0.4)\times 427

T_C=256.2^{\circ}C

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

3 0
3 years ago
HELP QUICK WILL GIVE BRAINLEST!!!!
Romashka-Z-Leto [24]
Hello!

The statement that <span>describes the solar feature shooting off into space labeled C is An envelope of plasma surrounding the Sun. 

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Have a nice day!</span>
3 0
2 years ago
Read 2 more answers
A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of
LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

\Psi=4RcE\pi

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\Psi =The energy fluency

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r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}

Then replacing our values we have that

I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}

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\Psi = 2.14*10^4 MeV/cm^2.s

The  uncollided flux density at the outer surface of the tank nearest the source is \Psi = 2.14*10^4 MeV/cm^2.s

6 0
2 years ago
How are the electric field lines around a positive charge affected when a second positive charge is near it? The field lines com
timofeeve [1]
The answer would be D hope it helps and sorry if it is wrong.  :)
3 0
3 years ago
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